String explanation

[Bernard Sigmund Gustav]

i just need some explanation why this is so:

i have:
assuming that var1 and var2 are equal

String somestring;
if(var1==var2)
   somestring = "something";

which works just fine

but if i do this

String somestring;
if(var1.equals(var2))
   somestring = "something";

it outputs an error that somestring has to be initialized so i have to:

String somestring=null;
if(var1.equals(var2))
   somestring = "something";

for this to work
why is that?

[Kia Phia Ben]

I think you code
String somestring ;

if(var1.equals(var2))
somestring = "something";
error because you not initialized value of variable somestring

[Bernard Sigmund Gustav]

if i use .equals and i don;t initialize somestring, it outputs an error
but if i use ==, it runs fine even if i don;t initialize somestring

[Garrett Rowe]

Maybe you could post your whole test code. I can't seem to duplicate the results you're getting.

If I try

public class InitTest
{
   public static void main(String[] args)
   {
      String x = "1";
      String y = "1";
 
      String aString;
 
      if (x == y) aString = "equal";
 
   }
}

or

public class InitTest
{
   public static void main(String[] args)
   {
      String x = "1";
      String y = "1";
 
      String aString;
 
      if (x.equals(y)) aString = "equal";
 
   }
}

They both compile fine. However if I try to use the String like:

public class InitTest
{
   public static void main(String[] args)
   {
      String x = "1";
      String y = "1";
 
      String aString;
 
      if (x.equals(y)) aString = "equal";
 
      System.out.println(aString);
 
   }
}

or I test identity (x == y)

public class InitTest
{
   public static void main(String[] args)
   {
      String x = "1";
      String y = "1";
 
      String aString;
 
      if (x == y) aString = "equal";
 
      System.out.println(aString);
 
   }
}

I get the same error, aString may not have been initialized.
[ March 10, 2006: Message edited by: Garrett Rowe ]

[Bernard Sigmund Gustav]

that's exactly my point
i did the one you did. it outputs the error
"may not be initialized", if i use .equals but if i just use ==, it prints just fine, even without initializing.

Originally posted by Garrett Rowe:
Maybe you could post your whole test code. I can't seem to duplicate the results you're getting.

public class InitTest
{
   public static void main(String[] args)
   {
      String x = "1";
      String y = "1";
 
      String aString;
 
      if (x.equals(y)) aString = "equal";
 
      System.out.println(aString);
 
   }
}

or I test identity (x == y)

public class InitTest
{
   public static void main(String[] args)
   {
      String x = "1";
      String y = "1";
 
      String aString;
 
      if (x == y) aString = "equal";
 
      System.out.println(aString);
 
   }
}

I get the same error, aString may not have been initialized.

[ March 10, 2006: Message edited by: Garrett Rowe ]

[Garrett Rowe]

My point was, this generates an error on my machine

public class InitTest
{
   public static void main(String[] args)
   {
      String x = "1";
      String y = "1";
 
      String aString;
 
      if (x == y) aString = "equal";
 
      System.out.println(aString);
 
   }
}

[Bernard Sigmund Gustav]

this did not generate an error for me.

public class InitTest
{   
   public static void main(String[] args)
   {      
      String x = "1";      
      String y = "1";       
      String aString;       
      if (x == y) 
         aString = "equal";       
      System.out.println(aString);    
   }
}

however, this did

public class InitTest
{   
   public static void main(String[] args)
   {      
      String x = "1";      
      String y = "1";       
      String aString;       
      if (x.equals(y)) 
         aString = "equal";       
      System.out.println(aString);    
   }
}

[Keith Lynn]

Your first code generates an error on my machine.

Is it possible that you accidentally include a ; after the if statement?

The reason it is giving you the error is that variables of automatic duration or local variables have to be initialized before they are used. Since you assign the variable aString a value following an if where the condition may or may not be true, it's possible that the variable doesn't get initialized.

Notice that this code compiles without any problem since the condition of the if statement is known to be true at compile-time.

public class InitTest
{   
   public static void main(String[] args)
   {      
      String x = "1";      
      String y = "1";       
      String aString;       
      if (true) 
         aString = "equal";       
      System.out.println(aString);    
   }
}

[ March 10, 2006: Message edited by: Keith Lynn ]

[Bernard Sigmund Gustav]

Originally posted by Keith Lynn:

The reason it is giving you the error is that variables of automatic duration or local variables have to be initialized before they are used. Since you assign the variable aString a value following an if where the condition may or may not be true, it's possible that the variable doesn't get initialized.

[ March 10, 2006: Message edited by: Keith Lynn ]

if this is so,then why does it only outpur an error if i use if(x.equals(y)) and no error when i use (x==y)???

[Keith Lynn]

Like I said, on my machine and on Garrett's machine it does generate an error. I'm not sure why it doesn't generate an error when you compile it.

[Bernard Sigmund Gustav]

then why does it generate an error oy yours and garrett's machine and not on mine? is that like, a bug in jdk or soemthing?

[Ernest Friedman-Hill]

An agressively optimizing compiler could turn this:

      String x = "1";      
      String y = "1";       
      String aString;       
      if (x == y) 
         aString = "equal";    

into this:

     String aString = "equal";

at compile time. I am not sure that's legal according to the JLS, but it's certainly a valid optimization. I know Borland's IDE compiler has done similar kinds of nonsense in some contexts; it looks like Shuini's compiler might be doing that. So what are you compiling with, exactly (vendor, version...)

[Jim Yingst]

[EFH]: I am not sure that's legal according to the JLS,

I'm pretty sure it's not legal according to the JLS, but I'm not particularly surprised that some compiler might do it. If x and y were declated final it would be a different story. And Shuini - are you sure you haven't oversimplified the code somehow? E.g. by omitting a "final" declaration? If not, I echo EFH - I would really like to know what compiler this comes from.

[Bernard Sigmund Gustav]

my java version is 1.4.2 which i downloaded from Sun and i'm using jcreator le to compile.

[Jim Yingst]

Shuini, let's be as specific as possible here. The following code generates errors for me using JDK 1.3_15, 1.4.2_10, and 1.5.0_06:

public class Test {
    public static void main(String[] args) {
        System.out.println(System.getProperty("java.version"));
        String x = "1";
        String y = "1";
        String aString;
        if (x == y)
            aString = "equal";
        System.out.println(aString);
    }
}

If you can indeed get it to compile and run, can you please tell us the exact version number it prints?